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GL
University of California, Berkeley

# A charged line is an electric current which carries a net electric charge.

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##### Top Educators  ##### Christina K.

Rutgers, The State University of New Jersey

LB ##### Aspen F.

University of Sheffield

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here will consider the electric field generated by a charged line. So here we have a piece of wire with length. Allen Charge Q. And so are linear charge density is simply the charge over the length. Q. Over Al, we want to know the electric field that's generated here at Point P and Point P is a distance perpendicular distance X from the center of this wire. Well, we know from superposition we can cut this wire into pieces. And let's say that this wire has a piece, has a length of D Y, and it's located a distance. Why above the center, which makes this distance here x squared plus y squared square root. Well, we know we can cut a piece on the opposite side at negative Y. And if we look at the fuel contributions from these, the bottom piece will give us a field in that direction. The top piece will give us a feel in that direction right away. By symmetry, we noticed that the white components cancel each other out, and so all we're left with is the X component. And so our total electric field E, as a function of X is simply some of each. Littlefield's X component and that is okay. Time some dick over the distance, which we know is X squared distance squared, which is X squared plus y squared times, cosine of data. And we know that close enough data is X over the square root of X squared plus y squared. So this becomes K X dick over X squared plus y squared to the three halves. And I pulled out KNX because their constant in any particular set point. So next we need to figure out like, Well, Dick simply our linear charge density Times are length of our segment, which is D y. And so Ari becomes que Lambda X, where I pulled out the Lambda from the integral cause. It's constant over de y times d y over x squared plus y squared to the three halves. And I should note, of course, this is in the extraction. Since this is a vector and also here, I can substitute our expression for lambda just k. Well, give us cake you x over l. I went to integrating de y we're X squared plus y squared to the three halves in the X direction and this integral will go from negative l over to to positive all over to again. I can note the symmetry of the problem and see that the bottom half here will contribute. The same amount of force is the top half. So I can simplify my integral by doubling it and just integrating over one half. And it doesn't matter which half I use. But why deal with negative signs If we don't have thio and this integral well, give us the value of El over X squared times the square root of l squared plus four x squared. So our total electric field once we simplify it as a function of X, it's too que que over x l squared plus four x squared and it's in the X direction. Now again, we'll note what happens if we squeeze this line into a point. In other words, we let l here go to zero. This cell here, well, the expression simply becomes too que que over x times The square root of four x squared, which gives us cake over X squared, which we expect from a point charge now in solving for the electric field, created a point B by this charge line, we argued that there was no white component. We based this argument on geometric considerations of the problem. But we can also do so mathematically. So let's consider the field generated by one segment of the line and more specifically, the segment the component that the segment creates in the Y direction. Well, this is simply the full magnitude of the field created by that segment times sign of data. And then we know that this is okay, Dick over X squared, plus y squared time sign of data, which is now why or the square root of X squared plus y squared. We know our dick from before is simply okay. Cue overall de y times is why all over x squared plus y squared to the three halves. Now we set up our integral for the full. Why component from the entire life, we're gonna pull out all of the constants like so and integrate Why d y over x squared plus y squared to the three halves and this is done from negative all over to toe all over to I can work this integral out. I can look it up in an integral table. However, one thing I will know is this inter ground Here is an R function. And if you're not, if you don't remember what not function it it simply means that if I let y equal why I get one value and if I let why equal negative why I get the exact same value but with a negative sign. And when you integrate an odd function over a symmetric region like this, the integral is just zero. And so we see mathematically that the white component of the electric field generated by this line at this particular point is zero. Now we can consider a special case of a charge line, a case in which the line is infinitely long. It has a charge per unit length of Lambda, as noted here, and we want to find the electric field at some point p a distance x away from the line now for the y component. We might have the same intuitions before where we think that there will be no white component because everything above this point, we'll cancel out everything below that point. But let's prove it mathematically. So we're gonna take a little cut here of some thickness D y at some height. Why? Well, the electric field in the Y direction from that segment, simply the electric field. Total electric field from that segment times sign of data. And we know when electric field will be k times the charge in that segment some tiny charged que over its distance, which we know here is X squared plus y squared the distance squared. I'm sign of data, which gives us why over X squared plus y squared square root. Remember our D? Q. Is the charge in that segment, which is just the charge per unit length, times the length, which is D y. And so we see that from this one component we get on electric field in the Y direction of K Y Lambda d Y over X squared plus y squared to the three halves, and so are totally y from the entire infinite line. It's simply going to be an integral We pull the cops in terms out. It's in a roll, go from negative infinity to infinity and we get why over X squared plus y squared to the three halves taken with respect to why again, as we did before. We'll note this in a Graham is an odd function. This is why it's one value, and it's when it's negative, why it's the same value, but with a negative sign. So this entire integral here is simply zero. So this line as a white component of the electric field of zero. So now we can calculate the X components that each segment generates, which is just the full field generated by the Component Times Co sign of data. And our d E is simply que de que, which we still have up here. Lambda D y over the distance squared, which is X squared plus y squared and casino theater, gives his ex over radical X squared plus y squared and simplifying that we get k lambda x d y over X squared plus y squared to the three halves. So the entire field generated by this infinite line is simply the integral, and I'll pull the constants out negative infinity to infinity de y over X squared plus y squared to the three halves. We'll note that this in a grand is positive, is even. I'm sorry and I even we mean that If we let y equal why we get one value and if we let y equal negative why we get the exact same value and we integrate even immigrant over a symmetric interval, we can simply double it and integrate over either half of the integral. And I'm sorry, that should be infinity. Not negative. Infinity. Yeah, we'll see this Integral is simply one over X squared. And so the electric field in the X direction, which is the entire electric field generated by this infinite line. It's simply to K lambda X over X squared or simplifying two K lambda over X. Now we might want to do a sanity. Check it the second. Do we have the right units? Well, to his unit lists, Que has units of Newton's Times meter squared per Coolum squared? And if you ever forget that, just remember Simple Equation Force, which is in Newton's. It's equal to K Q one Q two over R squared, and if you substitute cool longs for these and meter squared for R squared, you'll see these air. The correct units. Lambda is a charge per unit length, so columns per meter and X, of course, is the distance. So that's meters. So we see all of our meters. Cancel. We cancel one. Cool. I'm out. And so we're left with units of Newtons per Coolum. Exactly. The units we expect for an electric field.

GL
University of California, Berkeley
##### Top Physics 102 Electricity and Magnetism Educators  ##### Christina K.

Rutgers, The State University of New Jersey

LB ##### Aspen F.

University of Sheffield